Let \(Y\)  be a maximal subgroup of \(X\)  containing \(X_{x}\) . Then \(Y\)  is a large
subgroup of \(X\)  such that \(\left( \Phi _{ef}^{\ast }(p),\left|Y\right|\right) >1\)  by Theorem REF . Assume that \(Y\) 
is a member of \(\mathcal {S}(X)\) . Then the pairs \((X,Y)\)  are listed in [1]}, Proposition 4.28. Filtering such pairs \((X,Y)\)  with respect to the property \(\left( \Phi _{ef}^{\ast }(p),\left|Y\right|\right) >1\) , one of the
followings occurs:

\(X\cong P\Omega _{8}^{+}(q)\)  and \(Y\)  is isomorphic either to \(\Omega _{7}(q)\)  or to \(Sp_{6}(q)\)  according to whether \(q\)  is odd or even
respectively.

\(X\cong \Omega _{7}(q)\)  and \(Y\cong G_{2}(q)\) .

\(X\cong Sp_{6}(q)\) , \(q\)  even, and \(Y\cong G_{2}(q)\) .

\(X\cong PSp_{4}(q)\) , \(q\)  an odd power of 2, \(q>2\) , and \(Y\cong Sz(q)\) .

\((X,Y)\)  are listed in the Table REF .

<TABLE>Assume that Cases (1)–(4) hold. Since \(X_{x}\)  is a large subgroup of \(X\) ,
it follows that \(\left|X_{x}\right|>\left|X\right|^{1/3}\) . By using the lower bounds for \(\left|X\right|\)  provided in Lemma 4.3 of [1]}, we see that \(\left|X_{x}\right|\)  is greater than \(q^{28/3}/2\) ,
\(q^{7}/\@root 3 \of {4}\) , \(q^{7}/\@root 3 \of {2}\)  and \(q^{10/3}/\@root 3 \of {2}\)  in Cases
(1)–(4) respectively.
Assume that \(X\cong P\Omega _{8}^{+}(q)\) . Then \(Y\)  is isomorphic either to \(\Omega _{7}(q)\) , or to \(Sp_{6}(q)\) , according to whether \(q\)  is odd or even
respectively. Note that, \(q=2\)  is ruled out by Lemma REF . Also, \(q=3\)  cannot occur. Indeed, in this case \(13\mid \left|X\right|\) , \(13^{2}\nmid \left|X\right|\) , but 13 does not divide \(\left|X_{x}\right|\)  and \(k^{2}\) . Hence \(q>3\) .
Assume that \(X_{x} \ne Y\) . Then either \(X_{x}\cong \Omega _{6}^{-}(q).Z_{\epsilon }\) , where \(\epsilon =1,2\) , or \(X_{x}\cong G_{2}(q)\) , or \(X_{x} \cong Sp_{2}(q^{3}) \cdot Z_{3}\)  with \(q\)  even and \(2<q<2^{8}\)  by [3]}, Tables 8.28–8.30, 8.33–8.34 and 8.39–8.42, since \(\left|X_{x}\right|>\)  \(q^{28/3}/2\)  and \(\left( \Phi _{6f}^{\ast }(p),\left|X_{x}\right|\right) >1\) . It is not difficult to check that, \(k^{2}=[X:X_{x}]\)  has no solutions for \(X_{x} \cong Sp_{2}(q^{3}) \cdot Z_{3}\)  with \(q\)  even and \(2<q<2^{8}\) . So, this case is ruled out.
If \(X_{x}\cong \Omega _{6}^{-}(q).Z_{\epsilon }\) , then
\(k^{2}=\frac{q^{6}(q+1)(q^{2}+1)(q-1)^{2}(q^{2}+q+1)}{(4,q^{4}-1)\epsilon }\) 
and hence \(q^{2}+q+1\)  is a square, but this is impossible by [4]},
A7.1.
If \(X_{x}\cong G_{2}(q)\) , then \(k^{2}=\frac{q^{6}(q^{4}-1)^{2}}{(4,q^{4}-1)}\) 
and hence \(k=\frac{q^{3}(q^{4}-1)}{(2,q-1)}\) . Therefore \(k+1=q^{3}(q^{2}-1)\frac{q^{2}+1}{(2,q-1)}+1\) . Set \(\theta =(k+1,6(4,q^{4}-1)f)\) . Then \(\frac{k+1}{\theta }\mid \left|X_{x}\right|\)  by Lemma REF .
Furthermore, \(\theta =(k+1,3f)\) , since \(k+1\)  is odd. Thus \(\frac{k+1}{\theta }\mid \frac{q^{6}-1}{q-1}\) , since \(\frac{k+1}{\theta }\)  is coprime to \(q^{2}(q^{2}-1)^{2}\) . Then \(q^{3}(q^{2}-1)\frac{q^{2}+1}{(2,q-1)}<\frac{q^{6}-1}{q-1}\cdot 3f\) , which has no solutions for \(q>3\) .
Finally, assume that \(X_{x}=Y\) . We treat \(X_{x} \cong \Omega _{7}(q)\)  and \(X_{x} \cong Sp_{6}(q)\)  at the same time. Then \( k^{2}=q^{3 }(q^{2}-1)\frac{(q^{2}+1)}{(2,q-1)}\) , and since \(\frac{(q^{2}+1)}{(2,q-1)}\)  is odd, it follows that \(q^{2}-1\)  is a square. However, this is impossible by [4]}, A3.1.
Assume that \(X\)  is either \(\Omega _{7}(q)\) , or \(Sp_{6}(q)\)  for \(q\)  even, and
\(Y\cong G_{2}(q)\) . Assume that \(X_{x}\ne Y\) . Since \(\left|X_{x}\right|>q^{7}/\@root 3 \of {2}\)  and \(\left( \Phi _{6f}^{\ast }(p),\left|X_{x}\right|\right) >1\) , it follows from [3]},
Tables 8.30 and 8.41–8.43, that \(X_{x}\cong SU_{3}(q):Z_{\epsilon }\) , where
\(\epsilon =1,2\) , or \(q=3^{2m+1}\) , \(m\geqslant 1\) , and \(X_{x}\cong \)  \(^{2}G_{2}(q)\) . Thus,
\(k^{2}=\frac{q^{6}(q^{2}-1)^{2}(q^{2}+q+1)(q^{2}+1)}{(2,q-1)\theta }\text{,}\) 
where either \(\theta =(q+1)\epsilon \) , or \(\theta =1\)  respectively. So, \(q^{2}+q+1\) 
is a square in both cases, but this is impossible by [4]}, A7.1. Thus \(X_{x}=Y\)  and hence \(k^{2}=\frac{q^{3}(q^{4}-1)}{(2,q-1)}\) , which, as we have already seen, does not have solutions.
Assume that \(X\cong PSp_{4}(q)\)  and \(Y\cong Sz(q)\) , where \(q\)  is an odd
power of 2, \(q>2\) . Since \(\left|X_{x}\right|>q^{10/3}/\@root 3 \of {2}\)  and \(\left( \Phi _{4f}^{\ast }(p),\left|X_{x}\right|\right) >1\) ,
it follows that \(X_{x}=Y\)  by [3]}, Table 8.16. Hence \(k^{2}=q^{2}\left( q-1\right) \left( q+1\right) ^{2}\) , but this contradicts
[4]}, A3.1, since \(q\)  is an odd power of 2 and \(q>2\) . So far, we
have ruled out Cases (1)–(4).
Assume that Case (5) holds. Then \((X,Y)\)  are listed in Table REF , in which the symbol \(s\)  in Column 6, when recorded, denotes the largest
prime divisor of \(\left|X\right|\)  such that \(s\nmid \Phi _{ef}^{\ast }(p)\) , \(s \nmid \left|Y\right|\)  and \(s^{2}\nmid \left|X\right|\) . The previous
constraints imply \(s\mid \left[ X:X_{x}\right] \)  and hence \(s^{2}\mid \left|X\right|\) , which is a contradiction. Hence, only the groups,
whose corresponding cell in Column 6 of Table REF  is empty are
admissible, and these are in Lines 2 and 6–7 of the same table. The group \(X\cong PSL_{4}(2)\cong A_{8}\)  is ruled out in Lemma 3.5 of [10]}. If \(X\cong PSp_{4}(7)\)  and \(X_{x}\cong A_{7}\) , then \(7^{3}\)  is the highest power
of 7 dividing \(\left[ X:X_{x}\right] \) , a contradiction. Finally, if \(X\cong PSU_{4}(3)\)  and \(X_{x}\cong A_{7},PSL_{3}(4)\) , then \(k^{2}=1296\)  or 162. The latter is ruled out since it is not a square, whereas the former
yields \(k=36\) . Then \(k+1=37\)  must divide the order of \(G\)  by Lemma REF (3). However, this is impossible since \(G\leqslant P\Gamma U_{4}(3)\) , and the order of \(P\Gamma U_{4}(3)\)  has not such
a divisor. This completes the proof.
